# Two ways of evaluating $\int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}}$, giving discrepancy of a factor of $2$.

This is related to my previous question here. Here I am evaluating (in the sense of distributions) for a vector $$\mathbf{r} = (x,y,z)$$ $$\int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}} = \delta(x) \delta(y) \delta(z) \ .$$ At the same time, you can evaluate the above integral as $$\int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}} = \frac{1}{2 \pi^2 |\mathbf{r}|} \int_0^\infty dP \; P \sin(|\mathbf{r}| P) = - \frac{\delta'(|\mathbf{r}|)}{2\pi|\mathbf{r}|}$$ where the last line uses the representation $$\int_0^\infty \frac{dP}{\pi} P \sin( \alpha P) = - \delta'(\alpha)$$ (which follows from $$\int_0^\infty \frac{dP}{\pi} \cos( \alpha P) = \delta(\alpha)$$). There is additionally the identity $$\alpha \delta'(\alpha) = - \delta(\alpha)$$ would imply that $$\cdots = \frac{\delta(|\mathbf{r}|)}{2\pi|\mathbf{r}|^2} = 2 \delta(x)\delta(y)\delta(z)$$ where in the last line I have used the identity $$\delta(x) \delta(y) \delta(z) = \frac{\delta(|\mathbf{r}|)}{4\pi|\mathbf{r}|^2}$$ which is equation (3.1.30) from Kanwal's book on distributions.

The above seems to say that $$\delta(x)\delta(y) \delta(z) = 2 \delta(x)\delta(y) \delta(z)$$, which is obviously wrong. Where am I making a mistake in my second way of computing?

EDIT: Note that the $$3D$$ Fourier transform of a rotationally invariant function $$f(|\mathbf{p}|)$$ is also rotationally invariant $$F(|\mathbf{x}|)$$, $$F(|\mathbf{x}|) = \int \frac{d^3 \mathbf{p}}{(2\pi)^3} f(|\mathbf{p}|) e^{i \mathbf{r} \cdot \mathbf{p}} = \int \frac{d^3 \mathbf{p}}{(2\pi)^3} f(|\mathbf{p}|) e^{i |\mathbf{r}| |\mathbf{p}| \cos\theta }$$ where I have taken $$z \to |\mathbf{r}|$$ (free to do so by rotationally symmetry) and used spherical coordinates. From here the $$\phi$$ integral evaluates to $$2\pi$$, then doing the $$\theta$$ integral gives the $$\propto \sin(|\mathbf{r}|P)/|\mathbf{r}|$$ function in my formula above. I then use $$f(|\mathbf{p}|) =1$$ in the above.

• Could you please show how you do this step? $$\int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}} = \frac{1}{2 \pi^2 |\mathbf{r}|} \int_0^\infty dP \; P \sin(|\mathbf{r}| P)$$ Dec 9, 2020 at 17:27
• @md2perpe Sure thing, I added an edit in the above. This is a standard result for rotationally invariant 3D Fourier transforms Dec 9, 2020 at 18:27

Your computation of the integral using spherical coordinates $$\int \frac{d^3 \mathbf{p}}{(2\pi)^3} e^{i \mathbf{r} \cdot \mathbf{p}}=\frac1{(2\pi)^2}\int_{\Bbb R^+}\int_{[0,\pi]} r^2\sin\theta\cdot e^{i|{\bf r}|P\cos\theta}\,d\theta\,dP=\frac{\delta(|{\bf r}|)}{2\pi|{\bf r}|^2}$$ is correct as are the identities you have used up to this point. However, it is important to note the lack of consensus in the proportionality constant of $$|{\bf r}|^{-2}\delta(|{\bf r}|)$$. For example,
• Kanwal (2004) defines $$\delta(x) \delta(y) \delta(z) = \dfrac{\delta(|\mathbf{r}|)}{4\pi|\mathbf{r}|^2}$$ as you have stated, whereas
• Bracewell (1999) defines $$^3\delta(x,y,z)=\dfrac{\delta(\rho)}{2\pi\rho^2}$$ where $$\rho=|{\bf r}|$$ (see equation $$(51)$$ of this link).
This primarily stems from the ill-defined nature of $$x^{-\kappa}\delta(x)$$ when $$\kappa\ge1$$ as discussed in Delta function at the origin in polar coordinates and Delta function of the Euclidean norm $\delta(|\mathbf x|)$ / in polar coordinates at origin $\delta(r)$ for polar coordinates representations.